Question

# A man of mass M=50 kg stands at one end of a boat (of length L) which is floating in still water. The man walks to the other end of the boat. Length of the boat is L=10 m and mass M=20 kg. Then the distance moved by the boat relative to the ground is

A
507 m
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B
7.5 m
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C
257m
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D
12.57m
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Solution

## The correct option is A 507 mHere, no external force is acting on the system. Hence Xcom will not change its position. Assume that when the man reaches the other end, boat moves x distance left, relative to the ground. Hence, distance travelled by the man with respect to the ground, x1=(L−x)=(10−x) m Distance travelled by the boat with respect to the ground (x2)=−x m As we know, Xcom=m1x1+m2x2m1+m2 Here Xcom=0 ⇒0=50×(10−x)+(20)(−x)50+20 ⇒500−50x−20x=0 ⇒70x=500 ∴x=507 m towards left

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